A company hires management trainees for entry level sales positions. Past experience indicates that only 10% will still be employed at the e

Question

A company hires management trainees for entry level sales positions. Past experience indicates that only 10% will still be employed at the end of nine months. Assume the company recently hired six trainees.

a. What is the probability that three of the trainees will still be employed at the end of nine months?
b. What is the probability that at least two of the trainees will still be employed at the end of nine months?

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Faith 4 weeks 2021-11-09T00:49:22+00:00 1 Answer 0 views 0

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    2021-11-09T00:51:20+00:00

    Answer:

    a) 1.46% probability that three of the trainees will still be employed at the end of nine months.

    b) 11.43% probability that at least two of the trainees will still be employed at the end of nine months

    Step-by-step explanation:

    For each trainee, there are only two possible outcomes. Either they will still be employed at the end of nine months, or they will not. The probability of a trainer being employed is independent from other trainees. So we use the binomial probability distribution to solve this question.

    Binomial probability distribution

    The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    In which C_{n,x} is the number of different combinations of x objects from a set of n elements, given by the following formula.

    C_{n,x} = \frac{n!}{x!(n-x)!}

    And p is the probability of X happening.

    10% will still be employed at the end of nine months.

    This means that p = 0.1

    Assume the company recently hired six trainees.

    This means that n = 6

    a. What is the probability that three of the trainees will still be employed at the end of nine months?

    This is P(X = 3).

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    P(X = 3) = C_{6,3}.(0.1)^{3}.(0.9)^{3} = 0.0146

    1.46% probability that three of the trainees will still be employed at the end of nine months.

    b. What is the probability that at least two of the trainees will still be employed at the end of nine months?

    Either less than two will be employed, or at least two will. The sum of the probabilities of these events is decimal 1. So

    P(X < 2) + P(X \geq 2) = 1

    We want P(X \geq 2)

    So

    P(X \geq 2) = 1 - P(X < 2)

    In which

    P(X < 2) = P(X = 0) + P(X = 1)

    P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}

    P(X = 0) = C_{6,0}.(0.1)^{0}.(0.9)^{6} = 0.5314

    P(X = 1) = C_{6,1}.(0.1)^{1}.(0.9)^{5} = 0.3543

    P(X < 2) = P(X = 0) + P(X = 1) = 0.5314 + 0.3543 = 0.8857

    P(X \geq 2) = 1 - P(X < 2) = 1 - 0.8857 = 0.1143

    11.43% probability that at least two of the trainees will still be employed at the end of nine months

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